Problem: $f(x)=x-5$ $g(x)=\dfrac{x^2-12}{x^2+4}$ Write $g(f(x))$ as an expression in terms of $x$. $g(f(x))=$
Let's write $f(x)$ as the input to function $g$. $g({f(x)})=\dfrac{({f(x)})^2-12}{({f(x)})^2+4}$ Since $f(x)=x-5$, this becomes: $\begin{aligned} g({f(x)})&=\dfrac{({x-5})^2-12}{({x-5})^2+4}\\ \\ &=\dfrac{x^2-10x+25-12}{x^2-10x+25+4}\\ \\ &=\dfrac{x^2-10x+13}{x^2-10x+29}\\ \\ \end{aligned}$ Note: We simplified the result to obtain a nicer expression, but this is not necessary. The answer: $g(f(x))=\dfrac{x^2-10x+13}{x^2-10x+29}$